What are the solutions to the equation 0=290(8x-0.5x^2-1.4) ?

The solutions to the equation 0=290(8x-0.5x^2-1.4) are x=(40-3sqrt(170))/5 ,x=(40+3sqrt(170))/5

Find the zeros of 0=290(8x-0.5x^2-1.4)

The zeros of 0=290(8x-0.5x^2-1.4) are x=(40-3sqrt(170))/5 ,x=(40+3sqrt(170))/5

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0 = 290 (8 x - 0.5 x^{2} - 1.4)

x = \frac{40 - 3 170}{5}, x = \frac{40 + 3 170}{5}

Decimal Notation

x = 0.17695 \dots, x = 15.82304 \dots

Solution steps

0 = 290 (8 x - 0.5 x^{2} - 1.4)

Expand

290 (8 x - 0.5 x^{2} - 1.4) : 2320 x - 145 x^{2} - 406

0 = 2320 x - 145 x^{2} - 406

Switch sides

2320 x - 145 x^{2} - 406 = 0

Write in the standard form

a x^{2} + b x + c = 0

- 145 x^{2} + 2320 x - 406 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2320 \pm 232 0 ^{2} - 4 ( - 145 ) ( - 406 )}{2 ( - 145 )}

232 0^{2} - 4 (- 145) (- 406) = 174170

x_{1, 2} = \frac{- 2320 \pm 174 170}{2 ( - 145 )}

Separate the solutions

x_{1} = \frac{- 2320 + 174 170}{2 ( - 145 )}, x_{2} = \frac{- 2320 - 174 170}{2 ( - 145 )}

x = \frac{- 2320 + 174 170}{2 ( - 145 )} : \frac{40 - 3 170}{5}

x = \frac{- 2320 - 174 170}{2 ( - 145 )} : \frac{40 + 3 170}{5}

The solutions to the quadratic equation are:

x = \frac{40 - 3 170}{5}, x = \frac{40 + 3 170}{5}

2 (2 x - 2) \leq 2 x + 28

\frac{\frac{π}{2} + \frac{π}{4}}{2}

v + 1 > v - 6

y^{2 x} - y^{x} - 20

5 x^{2} + 2 x + 8 = 0

What are the solutions to the equation 0=290(8x-0.5x^2-1.4) ?
The solutions to the equation 0=290(8x-0.5x^2-1.4) are x=(40-3sqrt(170))/5 ,x=(40+3sqrt(170))/5
Find the zeros of 0=290(8x-0.5x^2-1.4)
The zeros of 0=290(8x-0.5x^2-1.4) are x=(40-3sqrt(170))/5 ,x=(40+3sqrt(170))/5